Is ${327753}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {327753}= &&{3}\cdot100000+ \\&&{2}\cdot10000+ \\&&{7}\cdot1000+ \\&&{7}\cdot100+ \\&&{5}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {327753}= &&{3}(99999+1)+ \\&&{2}(9999+1)+ \\&&{7}(999+1)+ \\&&{7}(99+1)+ \\&&{5}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {327753}= &&\gray{3\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {3}+{2}+{7}+{7}+{5}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${327753}$ is divisible by $9$ if ${ 3}+{2}+{7}+{7}+{5}+{3}$ is divisible by $9$ Add the digits of ${327753}$ $ {3}+{2}+{7}+{7}+{5}+{3} = {27} $ If ${27}$ is divisible by $9$ , then ${327753}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${327753}$ must also be divisible by $9$.